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Creating an Energy Efficient Engine

In a fuel cel...

Question

In a fuel cell, methanol is used as a fuel and $O_{2}$ is used as an oxidizer. The standard enthalpy of combustion of methanol is $- 726 {kJmol}^{- 1}$ . The standard free energies of formation of $C H_{3} O H (l)$ , $C O_{2} (g)$ and $H_{2} O (l)$ are − 166.3, −394.4 and $- 237.1 {kJmol}^{- 1}$ respectively. The efficiency of the fuel cell will be:

A

96.8 %

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B

66.2 %

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C

41.3 %

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D

85.1 %

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Solution

The correct option is A $96.8 %$
Points to remember:
1. $Δ_{r} G^{0} = Σ Δ_{f} G_{p r o d u c t s}^{0} - Σ Δ_{f} G_{r e a c t a n t s}^{0}$
2.Efficiency $% = \frac{u s e f u l w o r k}{i n p u t} \times 100$
$Δ_{r} G^{0} = u s e f u l w o r k$
3. $Δ_{f} G_{O_{2} (g)}^{0} = 0$
The reaction for the combustion of methanol is:
$C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
$Δ_{r} G^{0} = \frac{1}{2} Δ_{f} G_{C O_{2}}^{0} + 2 Δ_{f} G_{H_{2} O}^{0} - Δ_{f} G_{C H_{3} O H}^{0} - \frac{3}{2} Δ_{f} G_{O_{2}}^{0}$
On substituting values ,we get
$Δ_{r} G^{0} = - 702.6 k J m o l^{- 1}$

$e f f i c i e n c y = \frac{Δ G}{Δ H} \times 100 = 96.7 %$

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Similar questions

Q. If a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is

C H_{3} O H_{(l)} + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)

298 K

standard Gibb’s energies of formation for

C H_{3} O H (l), H_{2} O (l), and C O_{2} (g)

- 166.2, - 237.2

- 394.4 {kJ mol}^{- 1},

respectively. If the standard enthalpy of combustion of methanol is

- 726 {kJ mol}^{- 1}

, the efficiency of the fuel cell will be:

Q. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer.

The reaction is:

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

298 K

, standard Gibb's energies of formation for

C H_{3} O H (l), H_{2} O (l)

C O_{2} (g)

- 166.2, - 237.2

- 394.4 k J m o 1^{- 1}

If standard enthalpy of combustion of methanol is

- 726 k J m o 1^{- 1}

, efficiency of the fuel cell will be:

Q. In a fuel cell (the device used for producing electricity directly from a chemical reaction) methanol is used as a fuel and oxygen gas is used as an oxidizer. The standard enthalpy of combustion of methanol is

- 726 k J m o l^{- 1}

. The standard free energies of formation of

C H_{3} O H (l) . C O_{2 (g)}

H_{2} O (l)

- 166.3, - 394.4

- 237.1 k J m o l^{- 1}

respectively.
The standard free energy change of the reaction will be:

Q. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is:

{C H}_{3} O H (l) + \frac{3}{2} O_{2} (g) ⟶ \frac{1}{2} C O_{2} (g) + 2 H_{2} O (l)

Calculate standard Gibbs free energy change for the reaction that can be converted into electrical work. If the standard enthalpy of combustion for methanol is

- 726 k J {m o l}^{- 1}

, calculate the efficiency of conversion of Gibbs energy into useful work.

Δ G_{f}^{o}

{C O}_{2}, H_{2} O, {C H}_{3} O H, O_{2}

- 394.36; - 237.13; - 166.27

and zero respectively.

Q. Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for

H_{2} O (l), C O_{2} (g)

and pentane (g), respectively. The value of

E_{c e l l}^{0}

for the pentane-oxygen fuel cell is:

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